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Haus, Hermann A., and James R. Melcher. Solutions Manual for Electromagnetic
Fields and Energy. (Massachusetts Institute of Technology: MIT OpenCourseWare).
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Also available from Prentice-Hall: Englewood Cliffs, NJ, 1990. ISBN: 9780132489805.
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SOLUTIONS TO CHAPTER 10
10.0 INTRODUCTION
10.0.1
(a) The line integral of the electric field along 0 1 is from Faraday's law:
(1)
because no flux is linked (see Fig. S10.0.la). Therefore
-t/+iR = a
because the voltage drop across the resistor is iR. Hence
t/
= iR
(2)
R
v
v
c
+
F1sure BIO.O.la,h
The line integral along O2 is
4iR = d.A
dt
(3)
which leads to
(4)
1
Solutions to Chapter 10
10-2
Therefore, we find for the voltage across the voltmeter
1 dw.\
v=---
(5)
4 dt
(b) With the voltmeter connected to 2, (1) becomes
v = 2iR
Using (2),
and similarly for the other modes
. = 3 [1-dW.\]
v(3) = 3[IR]
­
4 dt
v(4) = 4iR = 4[! dW.\] = dw.\
4 dt
dt
For a transformer with a one turn secondary (see Fig. S10.0.lb),
v=
10.0.2
1 E· dl = !....
at
fa
!
B . da =
!!.w.\
dt
Given the following one-turn inductor (Figs. S10.0.2a and S10.0.2b), we want
to find (a) tI2 and (b) VI. The current per unit length (surface current) flowing
along the sheet is K = i/d. The tangential component of the magnetic field has to
have the discontinuity K. A magnetic field (the gradient of a Laplacian potential)
HIlS =
i
d
= 0
inside
(1)
outside
has the proper discontinuity. This is the field in a single turn "coil" of infinite width
d and finite K = i / d. It serves here as an approximation.
(a)
tI2
can be found by applying Faraday's law to the contour O2 ,
Using (I), and the constitutive relation B = PoD,
l
(B)
(A)a2
E·ds+
l(A)
d
E·ds=-(B)a2
dt
1
82
i(t)
Po-dxdy
d
(2)
Solutions to Chapter 10
10-3
Since the inductor waDs are perfectly conducting, E
on the left in (2). Therefore,
= 0 for the second integral
or,
slJjo diet)
d dt
~U2=---­
one-turn
inductor
K
= i(t)/d
~
d ~;""------'--.----"?I
-
/;/-__
y __
--
8
surface current, K,
flows through.
inductor walls
P,o~
I'"-----......'--+:z: - - - o f "
I
flows
through
this surface
----_ .. .. - ~
-~:::.=;.=----
Flsure 810.0.3
(b) Now, tl1 can be found by a similar method. Writing Faraday's law on 0 1 ,
(3)
Since 0 1 does not link any flux, (3) can be written
-til
d
= --(0)
=0
dt
Solutions to Chapter 10
10-4
10.1 MAGNETOQUASISTATIC ELECTRIC FIELDS IN
SYSTEMS OF PERFECT CONDUCTORS
10.1.1
The magnetic field intensity from Problem 8.4.1 is
i1fR2 [
1I( 1 1).
B = 4;2 cos 11 ,.s - b3 I r
. 1I( 1
+ sm
,.s + b23 ).16 ]
11
The E-field induced by Faraday's law has lines that link the dipole field and uniform
field. By symmetry they are tP-directed. Using the integral law of Faraday's law using
a spherical cap bounded by the contour r = constant, 9 = constant, we have
f
.
E· ds
= 21frsin9E~ = -
:t 1
6
J.'o B r 21frsin9rd9
di 1fR216
1
1
= -J.'o-21f~2sin9cos9d9(-- -3)
dt 41f 0
,.s b
di 1f R2
dt 4,..
(
1
1). 2
,.s - -b3 2sm 9
= -J.' ---1fr-2 o
Thus:
10.1.2
(a) The H-field is similar to that of Prob. 10.0.2 with K specified. It is z-directed
and uniform
=
inside
(1)
o outside
H. {K
Indeed, it is the gradient of a Laplacian potential and has the proper discon­
tinuity at the sheet.
(b) The particular solution does not need to satisfy all the boundary conditions.
Suppose we look for one that satisfies the boundary conditions at 11 = 0, Z = 0,
and 11 = a. IT we set
(2)
with Ezp(O, t) = 0 we have satisfied all three boundary conditions. Now, from
Faraday's law,
(3)
Integration gives
(4)
10-5
Solutions to Chapter 10
x=a
x=o
x=a
x=o
(b)
(a)
Figure
SlO.l.~a,b
The total field has to satisfy the boundary condition at y = -l. There, the field
has to vanish for almost all 0 ~ x ~ a, except for the short gap at the center of the
interval. Thus the E",-field must consist of a large field : E",p, over the gap 9, and
zero field elsewhere. The homogeneous solution must have an E",-field that looks
as shown in Fig. SlO.1.2a, or a potential that looks as shown in Fig. SlO.1.2b. The
homogeneous solution is derivable from a Laplacian potential cI>h
(5)
which obeys all the boundary conditions, except at y
cI>h at y = -l by
cI>h(y = -l) = aE",pf(x)
= -l.
Denote the potential
(6)
so that the jump of /(x) at x = a/2 is normalized to unity. Using the orthogonality
properties of the sine function, we have
- sinh ( m1l" l) ~ Am
a
2
= aE",p
fa
}",=o
/ (x) sin (m1l" x) dx
a
(7)
It is clear that all odd orders integrate to zero, only even order terms remain. For
an even order, except m = 0,
l
a
",=0
m1l"
/(x) sin (-x)
= 2
a
l
a 2
/
x
m1l"
- sin (-x)dx
",=0 a
a
mfr 2
/
2a
= -()2
usinudu
l
mll'
=
u=o
(~;)2 [ -
ucosul;;'fr/2
+
l
(8)
mfr 2
/ COSUdU]
= ~(_l)-'f+l
mll'
Therefore
m-even
m-odd
(9)
Solutions to Chapter 10
10-6
The total field is
dK {
~ 2(-1)m / 2 sinh!M
m1l"]
E = lJoi x[Y -I L.J
. h'::''/I' yI cos (-z)
dt
m
sma
a
•
-1)'1
m/2cosh ~'/I'y • (m1l" )]}
L.J 2(-1)
. h !Ml sm - z
m
sIn 2
a
~
10.1.3
(10)
~
... ~n
(a) The magnetic field is uniform and z-directed
B = i.K(t)
(b) The electric field is best analyzed in terms of a particular solution that satisfies
the boundary conditions at tP = 0 and tP = a, and a homogeneous solution
that obeys the last boundary condition at r = a,. The particular solution is tP­
directed and is identical with the field encircling an axially symmetric uniform
H-field
(1)
and thus
r
dK
E~ = -"2 IJ0 dt
(2)
The homogeneous solution is composed of the gradients of solutions to Laplace's
equation
(3)
At r = a, these solutions must cancel the field along the boundary, except at
and around tP = a/2. Because 8 < a, we approximate the field E</>h at r = a
as composed of a unit impulse function at tP = a/2 of content
a
dK
aE</>p = -"2alJ0dt
and a constant field
E</>h
(4)
dK
a
= "21J0dt
over the rest of the interval as shown in Fig. S10.1.3. Feom (3)
1 aCbh
1
E</>h r_G
_ = -a-atP
- = -a
I
L (n1l"/a) An cos (n1l"tP
-)
a
n
l_
T-E~p
Figure SI0.1.8
(5)
10-7
Solutions to Chapter 10
Here we take an alternative approach to that of 10.1.2. We do not have to worry
about the part of the field over 0 < ~ < a, excluding the unit impulse function,
because the line integral of E~ from ~ = 0 to ~ = a is assured to be zero (conser­
vative field). Thus we need solely to expand the unit impulse at ~ = a/2 in a series
of cos (~tr ~). By integrating
1
a
--(m7f/a)Am - = cos(m7f/2)aE<flp
a
2
(6)
where the right hand side is the integral through the unit impulse function. Thus,
(7)
Therefore
(8)
and
E= -
10.1.4.
~o
d: i{~
+
f:
2(_1)m/2(r/a)~-1
m_3
m-eYeD.
(9)
(a) The coil current produces an equivalent surface current K = Ni/d and hence,
because the coil is long
(1)
(b) The (semi-) conductor is cylindrical and uniform. Thus E must be axisym­
metric and, by symmet~, ~-directed. From Faraday's law applied to a circular
contour of radius r inside the coil
dB. 2
27fr E~ = - --7fr
dt
and
r
Ndi
E~ = -2~od dt
(c) The induced H-field is due to the circulating current density:
where we have set
i(t)
= I coswt
Solutions to Chapter 10
10-8
The H field will be axial, z- and ~independent, by symmetry. (The z- "inde­
pendence" follows from the fact that d::> b.) From Ampere's law
VxH=J
we have
dHz
--=J.,
dr
and thus
Hz induced
r2
N
= -wC1'4IL0"dlsinwt
For Hz induced <: Hz imposed for r ~ b
10.1.5
(a) From Faraday's law
a
at
and thus
VxEp=--B
(1)
aElIP
N di
--=-IL
-­
(2)
az
° ddt
Therefore,
(3)
(b) We must maintain E·n = 0 inside the material. Thus, adding the homogeneous
solution, a gradient of a scalar potential., we must leave E z = 0 at z = 0
and z = b. Further, we must eliminate E lI at y = 0 and y = a. We need an
infinite series
= An cos (~'Ir z) sinh (nb'lr y)
(4)
.h
L
n
with the electric field
At y =
±a/2
(6)
10-9
Solutions to Chapter 10
f(x)
=x-
~
-b/2
(a)
y
E"
E
Eta
Set - P. ~ ~ = p<lIIitive number
(b)
Flpre S10.1.1
We must expand the function shown in Fig. S10.1.5a into a cosine series. Thus,
multiplying (6) by cos ":,tI' z and integrating from z = 0 to z = b, we obtain
m1l"
b
Ndi
--A
cosh (m1l"
-·-a) = I/o b 2 m
2b
0 d dt
=
{
l
b
(
0
m,..
z - -b) cos -zdz
2
b
N IJi 2 ( b )2
-1/007
dt
m;r
(7)
m - odd
Om-even
Solving for Am
m - even
m- odd
The E-field is
E - _
N di
1/00 d dt
-
{(z _~)i
2
_ ~
'11
4bj(m'll")2
LJ cosh (m'll"aj2b)
n-odd
[sin (~,.. z) sinh (~'Il" y)lx
_ cos
(c) See Fig. SlO.1.5b.
(8)
(n;z) cosh (~'Il" Y)ly]}
(9)
Solutions to Chapter 10
10-10
10.2 NATURE OF FIELDS INDUCED IN FINITE
CONDUCTORS
10.2.1
The approximate resistance of the disk is
R= !211"a~
(J 2 at.
where we have taken half of the circumference as the length. The fiux through the
disk is [compare (10.2.15)1
A = J.' o i 2 a
2
This is caused by the current i 2 so the inductance of the disk L 22 is (using N = 1):
The time constant is
This is roughly the same as (10.2.17).
10.2.2
Live bone is fairly "wet" and hence conducting like the surrounding fiesh.
Current lines have to close on themselves. Thus, if one mounts a coil with its axis
perpendicular to the arm and centered with the arm as shown in Fig. 810.2.2, circu­
lating currents are set up. IT perfect symmetry prevailed and the bone were precisely
at center, then no current would fiow along its axis. However, such symmetry does
not exist and thus longitudinal currents are set up with the bone off center.
Flsure 810.2.2
10-11
Solutions to Chapter 10
10.2.3
The field of coil (1) is, according to (10.2.8)
(1)
The net field is
with Hind = K~ where K~ is the ¢J directed current in the shell. The E-field is
from Faraday's law, using symmetry
(2)
But
(3)
and thus, for r = a
2Hind
d
- + -Hind
=
/Aou!i.a
dt
d
--Ho
dt
(4)
In the sinusoidal steady state, using complex notation
(5)
and
(6)
where
/Aou!i.a
1"m= - ­
2
At small values of W1"m
(7)
10.3 DIFFUSION OF AXIAL MAGNETIC FIELDS THROUGH
THIN CONDUCTORS
10-12
'J
Solutions to Chapter 10
The circulating current K(t) produces an approximately uniform axial field
10.3.1
(1)
H. = K(t)
As the field varies with time, there is an induced E-field obeying Faraday's law
r
(2)
K= AuE
(3)
1 E.ds=-~
10
dt
18
#Lo B . da
The E-field drives the surface current
that must be constant along the circumference. Hence E must be constant. From
(1), (2), and (3)
K
d
2
4aE = 4a- = - _ I L Ka
(4)
Au
dt""o
and thus
d
4
-K+--K=O
dt
(5)
lJoUAa
Thus
(6)
with
1"m=
10.3.2
p-ouAa
-4-
(7)
(a) This problem is completely analogous to 10.3.1. One has
(1)
H. = K(t)
and, because K
be constant
= AuE must be constant along the surface, so that
(2d + V2d)E
Therefore
~
d
d2
d
dt
d
2
= --dt #LoK(t)2
K
Au
(2 + v2)- = --(lJoK)or
dK
K
-+-=0
dt
1"m
with
#Lo uAd
1"m
= 2(2 + V2)
E must
(2)
(3)
(4)
(5)
10-13
Solutions to Chapter 10
The solution for J =
K/ li. is
(6)
(b) Since
1
10
E.ds=O
(7)
1
and the line integral along the surface is V2dE, we have
(8)
(9)
(c) Again from Faraday's law
(10)
10.3.3
(a) We set up the boundary conditions for the three uniform axial fields, in the
regions r < b, b < r < a, r > a (see Fig. S10.3.3).
Ho(t) - H1 (t) = -Kout(t) =
-Joutli.
=
-uEoutli.
H 1 (t) - H 2 (t) = -KID(t) = -JIDli. = -uEiDli.
1
positive
direction ~
of K
Fleure SI0.S.S
(1)
(2)
Solutions to Chapter 10
10-14
From the integral form of Faraday's law:
21l"aEout
= -11-0 dtd [H1(t))1l"(a 2 -
b2) + H2(t)d 2]
2
(3)
(4)
21l" bEin = -11-0 :t [H2(t)d ]
We can solve for E out and E in and substitute into (1) and (2)
(1/:1 [a 2 - b2 dHdt)
b2 dH2(t)]
_
Ho ()
t - H 1 (t ) - 11-0""2
a
dt + -;----;;u-
(5)
_
(1/:1b dH2 (t)
H 1 ()
t - H 2 (t ) - 1I-0------;;u2
(6)
We obtain from (6)
(7)
where
Tm
lI-o(1/:1b
==-­
2
From (5), after some rearrangement, we obtain:
=> ~
m
2 + ~ (~-~) dHdt) + H (t) = H (t)
~ dH
1
mba
dt
a dt
0
(8)
(b) We introduce complex notation
Ho
= Hm coswt =
Re {Hme;wt}
(9)
Similarly H 1 and H2 are replaced by H 1,2 = Re IH1,2e;wtj. We obtain two
equations for the two unknowns III and II2 :
-Ill + (1 + iWTm)II2 =
0
b.
. (a b)]
[1 + 1WTm b - ~ H1 + ~1WTmH2 = Hm
A
A
They can be solved in the usual way
fI
0
1
=
IHm
1+iwTm
~iWTm
I
= _ (1 + iWTm)Hm
LJet
lJet
H
2= 11 + WT,:t~LJet-~) J I= _ LJet
II
m
m
where LJet is the determinant.
LJet == -{ [1 + iWTm(i -
~)](l + iWTm) + iWTm~}
10-15
Solutions to Chapter 10
10.3.4
(a) To the left of the sheet (see Fig. 810.3.4),
B = Koi-.
(1)
B=Ki.
(2)
To the right of the sheet
Along the contour Gl , use Faraday's law
1
10
E. ds =
_!!.
II K-K o
It,
I
!J,.
rB· da
(3)
dt Js
1
(T
/
=
(To
----..:----,=
1 + a cos !'f
1;1
I
Figure SI0.a.4
Along the three perfectly conducting sides of the conductor E = O. In the sheet the
current K - K o is constant so that
V·J=O~V·(uE)=O
i
E· ds =
(K Ii.
- K o ) dy = -I-'oabdK
1/=0
01
K - Ko
li.u
l
b
o
(4)
dt
00
r (1 + a cos bY ) dy = -I-'oab""dt
dK
1f
JI/=o
(5)
(6)
The integral yields b and thus
(7)
From (7) we can find K as a function of time for a given Ko(t).
(b) The y-component of the electric field at :t: = -a has a uniform part and a
y-dependent part according to (5). The y-dependent part integrates to sero
and hence is part of a conservative field. The uniform part is
Ewb = -
K-Ko
Ii.
000
dK
b = I-'oabdt
(8)
Solutions to Chapter 10
10-16
This is the particular solution of Faraday's law
(9)
with the integral
dK
E yp = -I-'ozdi"
(10)
and indeed, at z = -a, we obtain (8). There remains
Ellh
= -
K - Ko
!:iu
o
acos
('lrY)
b
(11)
It is clear that this field can be found from the gradient of the Laplacian
potential
(12)
~ = A sin
sinh (~z)
<'7)
that satisfies the boundary conditions on the perfect conductors. At z = -a
8~
--
8y
I
'lrY.
'Ir
:1:=-4
<'Ira
= -Acos-smh - )
b
b
b
- K o acos'lry
=- K!:iu
b
(13)
o
and thus
(14)
10.4 DIFFUSION OF TRANSVERSE MAGNETIC FIELDS
THROUGH TmN CONDUCTORS
10.4.1
(a) Let us consider an expanded view of the conductor (Fig. 810.4.1). At y = !:i,
the boundary condition on the normal component of B gives
(1)
11
(a)
(e)
~
(IT,lL)
(b)
F1sure 910.4.1
10-17
Solutions to Chapter 10
Therefore
(2)
At y= 0
(3)
Since the thickness, 11, of the sheet is very small, we can assume that B is uniform
across the sheet so that,
(4)
Using (3) and (4) in (2),
BG-Bb=O
11
11
(5)
From the continuity condition associated with Ampere's law
Since
K
= K.I.,
n
_HG+Hb
III
III
= I,.,
=K
•
(6)
The current density J in the sheet is
J _ K.
•-
(7)
11
And so, from Ohm's law
E _ K.
• - l1a
(8)
Finally from Faraday's law
BD
VxE=-­
Bt
(9)
Since only B II matters (only time rate of change of flux normal to the sheet will
induce circulating E-fields) and E only has a z-component,
BE.
- Bz
BBII
=-lit
From (8) therefore,
and finally, from (6),
(10)
(b) At t = 0 we are given K
sheet, we have J = 0
= I.Kosinpz. Everywhere except within the current
=> B
= -V\If
Solutions to Chapter 10
10-18
So from V . ,",oH
= 0, we have
Boundary conditions are given by (5) and (10) and by the requirement that the
potential mut decay as y -+ ±oo. Since Hz will match the sinfJz dependence
of the current, pick solutions with cos fJz dependence
w(a)
= A(t) cos fJze- fJ1I
(l1a)
web) = O(t) cos fJze fJ1I
H(a)
(l1b)
= fJA(t) sin fJze- fJ1I i x + fJA(t) cos fJze- fJ1I i y
fJO(t) sin fJze fJ1I i x
H(b) =
-
fJO(t) cos fJzefJ1Iiy
(12a)
(12b)
From (5),
Therefore,
A(t)
= -O(t)
(13)
From (10),
:z [fJA(t) sin fJze-1J1I11I=0 - fJO(t) sin fJ ze {J1I1 1I=0]
dA(t)
= -dO',",ofJ cos fJze- fJ1I 111=0 dt"
Using (13)
2fJ2 A(t) cos fJz
dA(t)
= -dO',",ofJ cos fJzdt"
The cosines cancel and
dA(t)
dt
+ ~A(t) =
dO',",o
0
(14)
The solution is
A(t) = A(O)e- t / r
(15)
So the surface current, proportional to Hz according to (6), decays simila.rly
as
10-19
Solutions to Chapter 10
10.4.2
(a) IT the sheet acts like a perfect conductor (see Fig. S10.4.2), the component of
B perpendicular to the sheet must be sero.
y
y=d
)--~-c~~-{i()------z
IL
K(t)
--+ 00
Figure
At y
= 0 the magnetic field
= i.K(t) cos{jz
SlO.4.~
experiences a jump of the tangential component
(1)
with n
II i)'
and B 2
= 0,
Hz
= -K(t)cospz
(2)
The field in the space 0 < y < d is the gradient of a Laplacian potential
'ilf = A sin pz cosh P(y - d)
(3)
The cosh is chosen so that HIJ is sero at y = d:
B
= -AP[cospzcoshP(y -
d)i x
+ sin pz sinh P(y -
d)i)']
(4)
Satisfying the boundary. condition at y = 0
-ApcospzcoshPd = -K(t) cospz
Therefore
A=
K(t)
pcoshPd
'ilf = K(t) sinpzcoshP(y - d)
pcoshPd
(5)
(6)
(7)
(b) For K(t) slowly varying, the magnetic field diffuses straight through 80 the
sheet acts as if it were not there. The field "sees- IJ - 00 material and,
therefore, has no tangential H
'ilf = A sin pz sinh f3(y - d)
(8)
Solutions to Chapter 10
10-20
which satisfies the condition Hz = 0 at y = d. Indeed,
B
= -AP[cosp:r:sinhp(y -
d)i x
+ sinpzcoshP(y -
d)l~1
Matching the boundary condition at y = 0, we obtain
A= _
K(t)
P sinh Pd
q; = _ K(t) sin pzsinh Ply - d)
P sinh Pd
(9)
(10)
(c) Now solving for the general time dependence, we can use the previous results
as a clue. Initially, the sheet acts like a perfect conductor and the solution
(7) must apply. As t - 00, the sheet does not conduct, and the solution
(10) must apply. In between, we must have a transition between these two
solutions. Thus, postulate that the current 1.K, (t) cos pz is flowing in the top
sheet. We have
.
K,(t)cospz
= ut::..E.
(11)
Postulate the potential
.q; = O(t) sin pz cosh P(y - d) _ D(t) sinp:r:sinh P(y - d)
pcoshPd
psinhpd
(12)
The boundary condition at y = 0 is
8q;
- 8z 11/=0
= Hz 11/=0 = -K(t)cospz
(13)
= -O(t) cos pz - D(t) cos pz
Therefore
O+D=K
(14)
8q; 1
I
cospz
- 8z lI=d = Hz lI=d = K, (t) cos pz = -O(t) cosh Pd
(15)
At y= d
The current in the sheet is driven by the E-field induced by Faraday's law
and is z-directed by symmetry
8E. __
8y -
!...
8t IJo
H _
z - lJo
cospzcoshP(y - d) dO
cosh Pd
dt
cos p:r:sinh Ply - d) dD
- lJo
sinh Pd
dt
(16)
Therefore,
E _ lJo cos p:r:sinhp(y - d) dO
• -
pcoshPd
dt -lJo
cos pz cosh P(y - d) dD
psinhPd
dt
(17)
Solutions to Chapter 10
10-21
At y= d
Ez
=
dD
1
-1-'0 [3 sinh [3d cos [3xdI
=
K. cos [3x
u!J.
(18)
Hence, combining (14), (15), and (18)
cosh [3dK.
I-'ou!:::&
dD
= -C(t) = -K + D = --[3coth[3ddI
(19)
resulting in the differential equation
dD D
I-'ou!:::&
Rd -+
- c o th l'
= K
[3
dt
(20)
With K a step function
(21)
where
1"m
I-'ou!:::&
= - - coth [3d
[3
and
C
At t = 0, D = 0 and at t
obtained solutions.
10.4.3
=
(22)
= Koe-tlrm
00,
C
= O.
This checks with the previously
(a) If the shell (Fig. 810.4.3) is thin enough it acts as a surface of discontinuity
at which the usual boundary conditions are obeyed. From the continuity of
the normal component of B,
B ra
1If
(T
-
B rb = 0
t
o
~
(1)
Ho
(a)
(b)
Figure 810.4.3
Solutions to Chapter 10
10-22
the continuity condition associated with Ampere's law
(2)
use of Ohm's law
J
K
E=-=-
U
!1u
(3)
H: - Hg = Kif> = !1uEIf>
(4)
results in
The electric field obeys Faraday's law
aB
(5)
VxE=-­ at
Only flux normal to the shell induces E in the sheet. By symmetry, E is <p-directed
1 a(
.)
( v x E )r = -'-0 ao Elf> Sin 0 =
rSIn
aBr
--a
t
(6)
And thus, at the boundary
1
RsinO
aoa [.Sin O[H9
G
-
H b]
9
A
aHr
= -J.&ouu---;jt
(7)
(b) Set
Ho(t) = Re {Hoeiwt}[cosOi.. - sinOi9 ]
(8)
The H-field outside and inside the shell must be the gradient of a scalar
potential
.9.
Acos 0
Wa
= -HorcosO
+ -2
­
r
Wb = GrcosO
o
ii = -HosinO +
~ sinO
r
iig = GsinO
ii: = HocosO
ii~
2A
+ 3'"
cosO
r
(9)
(10)
(11)
(12)
(13)
= -GcosO
(14)
'"
= Brb ~ H o + 2A
R3 = -0
(15)
From (1)
a
Br
Introducing (11), (12), and (13) into (7) we find
1
RsinO
. 2 0( -Ho+ R3
1 -0"')} =
aoa { Sin
.
{HocosO+ 21R3
cos 0 }
-JWJ.&o!1u
(16)
10-23
Solutions to Chapter 10
from which we find A, using (15) to eliminate O.
.A =
_ iWIJot::..uR4 H o
2(iwIJot::..uR
+ 3)
(17)
.A provides the dipole term
m =.A =
411"
-iwIJot::..uR4Ho
2(iwIJ ot::..uR + 3)
(18)
and thus
(19)
with
IJout::..R
1'= :.....:...._­
3
(c) In the limit
WT -+ 00,
we find
as in Example 8.4.4.
10.4.4
(a) The field is that of a dipole of dipole moment m
= ia
ia
W= --cosO
2
411"r
(1)
(b) The normal component has to vanish on the shell. We add a uniform field
sa
W= Ar cos 0 + - - 2 cosO
411"r
(2)
The normal component of Hat r = R is
-aWl
-
ar r=R
=0= - (
Ai
-a
2 - - ) cosO
411"R3
and thus
(3)
and
(see Fig. SI0.4.4).
Solutions to Chapter 10
10-24
Flpre 810.4.4
(c) There is now also an outside field. For r < R
ia
qr = 411T2 cos 8 + A(t)r cos 8
(5)
qr = O(t) cos 8
(6)
For r > R,
r 2
The 8-components of B are
H(J =
4~:S sin(J + A sin (Jj
r<R
(7a)
and
r>R
(7b)
2ia
)
H r = ( -Rs - A cos(J
(8a)
H(J = 0 sin (Jj
,.s
The normal component at r = R is
41f
and
Hr
=
20
RS cosO
(8b)
With the boundary condition (7) of Prob. 10.4.3, we have
1
RsinO
a [ . 2 (0
ao Sin 0 RS -
ia
)]
2p. o l1u
dO
41fRs - A = ---w-cos 0 d;
(9)
From the continuity of the normal component of B, we find
(10)
10-25
Solutions to Chapter 10
The equation for 0 becomes
1
R4 sine
a [ . 2 e(o _
ae
sm
ia
411"
or
T:
m
with
'Tm
+
20 _ 2ia)] = _ 2po li.u
edO
411"
R3 cos dt
dO
ia
-+0=-
dt
411"
(11)
(12)
= Po uIi.R/3. IT we consider the steady state, then
[Cei"'tj
0= Re
C=
A=
(13)
ia
1
(14)
(1 + iw'Tm ) 411"
=
2ia _ 20
411"R3
R3
2ia
iW'Tm
411"R3 1 + ;W'Tm
(15)
Jointly with (5) and (6), this determines \li.
(d) When W'Tm -+ 00, we have C -+ 0, no outside field and A = 2ia/411"R3 which
checks with (3). When W'Tm -+ 0, we have no shield and A -+ O. The shell
behaves as if it were infinitely conducting in the limit W'Tm -+ 00.
10.4.5
(a) IT the current density varies so rapidly that the sheet is a perfect conductor,
then it imposes the boundary condition (see Fig. 810.4.5),
r=b
D'PoB=O at
.
"
., :
....
. .: .":
. .. .
. -.. .....
.
.",
. ..
:,'
·· ....
0.··
"
·'
K
= K(t) sin 2,pi•
.--:-.-'"7',- .~
'
•
0•••
..... ."
0"
':"
.
. . ..
"
•
'0'
",: . • f '.:'
•
:
'
. ' .'
.~
.
"
.'
~.'
.:
:
I
..
"'-
-
Figure 810.4.5
" ' :." ";. p.
-+ 00
:
..
Solutions to Chapter 10
10-26
Inside the high Il. material H = 0 to keep B finite. So at r = a,
nxH=K
Therefore
-i.H~ =
K(t) sin 24>i.
Thus, the potential has to obey the boundary conditions
8'iJ1
8r
-=0
at
_!8'iJ1 =-K(t)sin24>
r 84>
(1)
r=b
at
r=a
(2)
In order to satisfy (2), we must pick a cos 24> dependence for 'iJI. To satisfy (1), one
picks a [(r/b)2 + (b/r)2] cos 24> type solution. Guess
Indeed,
8'iJ1 =
a;:
A [2r
b2
~:
2
-
2b ] cos 24> = 0 at r = b
-;:3
= -A[(r/b)2 + (b/r)2J2sin24>
From (2),
~[(a/b)2
+ (b/a)2J2sin 24> = -K(t) sin 24>
a
Therefore,
'iJI _
K(t)a [(r/b)2 + (b/r)2J
24>
+ (b/a)2] cos
(3)
- - - 2 - [(a/b)2
(b) Now the current induced in the sheet is negligible, so all the field diffuses
straight through. The sheet behaves as if it were not there at all. But at r = b
we have J.' - co material, so H = 0 inside. Also, since now there is no K at
r = b, we must have
H~ = 0 at r = b
It is dear that the following potential obeys the boundary condition at r = b
'iJI = A[(r/b)2 - (b/r)2J cos 24>
H~ = _! 8'iJ1
r 84>
=
~[(r/b)2 r
(b/r)2]2sin24> = 0 at
Again, applying (2)
A [(a/b)2 _ (b/a)2J2 sin 24> = -K(t) sin 24>
a
r= b
10-27
Solutions to Chapter 10
Thus,
\11
=
K(t)a l(r/b)2 - (b/r)2]
2~
--2-I(a/b)2 - (b/a)2] cos
(4)
(c) At the sheet, the normal B is continuous assuming that l::.. is small Also, from
Faraday's law I
dB
VxE=-­
(5)
dt
Since only a time varying field normal to the sheet will induce currents, we
are only interested in (V X E)r
By symmetry there is only a z-component of E
1 aE
-;a~
_ aBr
(6)
• --at:
One should note, however, that there are some subtleties involve in the deter­
mination of the E-field. We do not attempt to match the boundary conditions
on the coil surface. Such matching would require the addition of the gradient
of a solution of Laplace's equation to E p = i.E•. Such a field would induce
surface charges in the conducting sheet, but otherwise not affect its current
distribution. Remember that in MQS Eo BE is ignored which means that the
charging currents responsible for the bUfCI-up of charge are negligible com­
pared to the MQS currents flowing in the systems.
Feom Ohm's law, J
= uE. But, J = K/l::...
1
a K.
aBr
-; a~ l::..u =-at
(7)
Applying the boundary conditions from Ampere's law,
n X IHgaplr=b -
H,.._oo]
= K.i.
So at r = b
(8)
Now guess a solution for \11 in the gap. Since we have two current sources (the
windings at r = a and the sheet at r = b) and we do not necessarily know
that they are in phase, we need to use superposition. This involves setting up
the field due to each of the two sources individually
Solutions to Chapter 10
10-28
Here, A represents the field due to the current at r = b, and G is produced
by the current at r = a. Apply the boundary condition (2), at r = a. We find
from the tangential H-field
2G(t) [(a/b)2 _ (b/a)2]
a
= -K(t)
Thus,
G(t)
-aK(t)
(b/a)2]
= 2[(a/b)2 -
(10)
The normal and tangential components of H at r = b are
Hr
2b
2a2
4
= -{A(t)[a2 + 63] + G(t),)COS2¢
H", =
{A~t) [(b/a)2 -
(a/b)2]}2sin2¢
(11)
(12)
From (8)
lSo~Ub :¢ [A~t) [(b/a)2-(a/b)2]2sin 2¢] = {(:: + 2;2) a~~t) +~ ~~} cos2¢
Using (10),
dA(t) A
2
[«(1/6)2 - (6/a)21
---;j,t + (t) lSobAu [(a/b)2 + (b/a)2]
a
dK(t)
= [(a/b)2 + (b/a)2][(a/b)2 - (b/a)2] dt
Simplifying,
aA(t) + A(t)
at
r
r
= DdK(t)
at
lJobAu [(a/b)2 + (b/a)2]
[(a/b)2 _ (b/a)2]
= - 2-
a
D = [(a/b)2
+ (b/a)211(a/b)2 _ (b/a)2]
(13)
(14)
(15)
dK/dt is a unit impulse function in time. The homogeneous solution for A is
A(t) ex e- t / r
(16)
and the solution that has the proper discontinuity at t = 0 is
A=DKo
(17)
10-29
Solutions to Chapter 10
'If _
-
-aKa [(r/b)2 + (b/r)2]
2.
2
(a/b)2 + (b/a)2 cos ~
It is the same as if the surface currents spontaneously arose to buck out the
field. At t -+ 00, e- t /.,. -+ 0
'If
-aKa [(r/b)2 - (b/r)2]
(a/b)2 _ (b/a)2 cos 2~
= - 2-
This is when the field has enough time to diffuse through the shell
if no surface currents were present.
10.4.6
80
it is as
(a) When w is very high, the sheet behaves as a perfect conductor, and (see Fig.
810.4.6)
,T. _ bK[(r/a) + (a/r)J
A.
(1)
'.I!' [b a] cos."
ii+;;
a ar = 0 at r = a, and -
Then, indeed, 'If/
K.
t ~ accounts for the surface current
K{t) = KD{t) sin 41
i
" .
,
"
'.p-+oo
...
...
' .• I
, '
., : ' .
0..
'. : .' .
... " .
• : '. :'
.~ .: : • :.::
',
..
Figure 810.4.6
0_
.
Solutions to Chapter 10
10-30
(b) When w is very low, then
a'JI/at/J =
,T. _
0 at r = a and
bK!(rla) - (aIr)]
[~_~}
'J!'-
A.
cos¥'
(2)
(c) As before in Prob. 10.4.5, we superimpose the field caused by the two current
distributions
'JI = {A(t)[ ~ -
~} + O(t)[!: - ~J} cos ~
a r b r
The r- and
~-components of
H r
(3)
the field are:
= -{ A(t)[~ + ~] + O(t)[~ +
:2]} cos~
H4> = {A(t) [!: _~] + O(t) [!: _ ~]}sin~
r a r
r b r
(4)
(5)
At r = b,
(6)
and thus
A(t)
= ~o(t~
(7)
ii-b
At r = a,
-H<t>lr=a = K.
where K. is the current in the sheet. From (7) of the preceding problem
solution, we have at r = a
1 a H",
aHr
- - - - - = -}Jo-
a a~
Au
at
(8)
Thus, using (4) and (5) in (8):
O(t) [~ _ !} = -}JoAua{ ~ dA(t) + dOlt) [! + ~)}
a b a
a dt
dt b a2
(9)
Replacing A through (7) we obtain
dO + [~- ~}O(t)
dt
l-'oAua[~ +~]
=
2b
dKo(t)
(a/b)2 - (6/a)2 dt
(10)
Thus
(11)
with
(12)
10-31
Solutions to Chapter 10
D=
2b
(alb)2 - (bla)2
The solution for a step of Ko(t) is
= DKoe-t/f'
C
(13)
o
-t/f'
C( t ) -- DKoe -t/f' -_ (alb)22bK
_ (bla)2 e
Combining all the expressions gives the final answer:
'Ii = ;C0b
-a - -ab
For very short times
{[!: -~] - [! +-]
- ~]
2
a
r
[-b
e-t/f'}cos tP
a
tiT <: 1, one has
which is the same as (1). For very long times exp -tiT = 0 and one obtains
(2).
10.5 MAGNETIC DIFFUSION LAWS
10.5.1
(a) We first list the five equations (10.5.1)-(10.5.5)
VxB=J
(10.5.1)
J =O'E
(10.5.2)
a
(10.5.3)
V·p.B=O
(10.5.4)
V·J=O
(10.5.5)
vxE=
- atp.B
Take the curl of (10.5.3) and use the identity
(1)
also note that
because
0'
v .J
= V . O'E = o'V . E = 0
(2)
is uniform. Therefore,
2
a
-V E=--Vxp.B
at
(3)
Solutions to Chapter 10
10-32
or
-V 2 (Jju) =
-IJ~J
at
(4)
(b) Since J = i.J., equation (b) follows immediately from (4). We now use
(10.5.S)
a
v x (Jju) = - atlJH
But
v x (J ju) = !uv x (i.J.(z, y)) = !u (i x a8y J. -
and thus
aH
a (J.).
at = - 8y
UIJ Ix
8
i,. a J.)
z
8 (J.).
+ 8z
UIJ I,.
10.6 MAGNETIC DIFFUSION TRANSIENT RESPONSE
10.6.1
10.6.2
The expressions for H. and JfJ obey the diffusion equation, no matter what
signs are assigned to the coefficients. The summations cancel the field -K"zjb and
current density K"jb respectively, at t = a and eventually decay. IT one turns off a
drive from a steady state, the current density is initially uniform, equal to K"jb and
the field is equal to -K"zjb and then decays. But, the symmations with reversed
signs have precisely that behavior.
(a) The magnetic field is
(1)
H=i.H. = K"
and there is no E-field, nor J within the block.
(b) When the current-source is suddenly turned off, the H-field cannot disappear
instantaneously; the current returns through the conducting block, but still
circulates in the perfect conductor around the block. For this boundary value
problem we must change the eigenfunctions. At z = 0, the field remains finite,
because there is a circulation current terminating it. Thus we have, instead
of (10.6.15),
00
H.=
I:
OnCOS(~:z)e-t/T"
(2)
n-odd
with the decay times
4IJub2
Tn
(S)
= (ml")2
Initially, H. is uniform, and thus, using orthogonality
m1l"
2b • m1l"
H. cos -zdz = K,,- sm -b
2b
m1l"
2
1
0
b
2
= -Om
(4)
10-33
Solutions to Chapter 10
and thus
m-I(
C". = (-1)"- - 4: )
K p ;
m",
. -I 4:
L: (-1)-'---K
R'"
H. =
cos (R"')
- : r ; e- tl r ..
p
n-odd
(5)
mood
26
The current density is
J1I
aHa = -1::(-1)
2
~
. (R"')
tl
= --, Kpsm
- : r ; e- r ..
a:r; b
2b
H we pick a new origin at
. (R"')
sm
-:r;
2b
:r;'
= :r; + b, then
. (R"',
R'" , sm
. (R"')
= sm
-:r; - -R"') = - cos -:r;
­
2b
2
2b
2
= -(-1) ~
:I
cos (R"')
-:r;'
2b
for
R
odd
Interestingly, we find
At t = 0 this is the expansion of a unit impulse function at :r:' = 0 of content
-2Kp • All the current now :Hows through a thin sheet at the end of the block.
The factor of 2 comes in because the problem has been solved as a SYMmetric
problem at :r:' = 0, and thus half of the current ":Hows· in the "imagined·
other half.
10.1 SKIN EFFECT
10.1.1
(a) In order to find the impedance, we need to know the voltage
current being k •. The voltage is (see Fig. 10.7.2)
.
tI,
the complex
(1)
and, from Faraday's law
(2)
From (2) and (10.7.10)
(3)
Solutions to Chapter 10
10-34
and thus the impedance is at :t = -b
(4)
But the factor in front is
iawp.o _ a(l + i)
d(1 + j) - duo
(5)
(b) When b <: 0, we can expand the exponentials and obtain
Z
= a(1 + i) 1 + (1 + iH + 1 - (1 + iH
duo 1 + (1 + iH - 1 + (1 + iH
a(1 + i)
1
a
= duo (1 + iH = dub
(6)
(c) When b ,. 0, then we need retain only the exponential exp[(1
the result:
z=
+ i)b/ol
a(l + j)
duo
so that
with
(7)
a
Re(Z) = ­
duo
This looks like (6) with b replaced by
10.7'.2
o.
(a) When the block is shorted, we have to add the two solutions exp±(1
so that they add at the termination. Indeed, if we set
+ i)f
(1)
then the E-:6.eld is, from
(2)
and thus through integration
(3)
and is indeed zero at z
A so that
= o. In order to obtain Hz = k.
at z
= -b we adjust
(4)
Solutions to Chapter 10
10·35
(b) The high frequency distribution is governed by the exp -(1 + i) i(:I: < 0) and
thus
II "'" k e-(1+" f = k -(1+i) £j!
(5)
II -
•
e(l+i) k
.e
6
This is the same expression as the one obtained from (10.7.10) by neglecting
exp -(1 + i)i and exp(1 + i)b/o.
(c) The impedance is obtained from (3) and (4)
aE
11
dK.
Iz=-b --
a(1 + i)
duo
e(1+i)b/6 - e-(1+i)b/6
-'-:-~. ~~-;:------;-::~=
e(1+i)b/6
+ e-(1+i)b/6